A 3.00 L flask contains 0.200 mol of an ideal gas at a pressure of 1.20 atm. What is the temperature of the gas in degrees Celsius? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
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Correct answer
A. -53.6 °C
Principle or equation
Ideal gas law: PV = nRT. Solve for T in kelvin, then convert to Celsius: T(°C) = T(K) - 273.15.
Why this answer is correct
T = PV / (nR) = (1.20 atm × 3.00 L) / (0.200 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 3.60 / 0.01642 = 219.2 K. Convert: 219.2 - 273.15 = -53.95 °C ≈ -53.6 °C.
Example
For 0.300 mol in 5.00 L at 1.40 atm, T = (1.40×5.00)/(0.300×0.0821) = 284.2 K = 11.0 °C.
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